Question

Please help!!!!!!!:>

Answer 1

`\boxed{\boxed{\sf{(x-3)(x+16)~or~(x+16)(x-3)}}}`

Solution Steps:

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1.) Factor the expression by grouping:

- First, the expression needs to be rewritten as
`x^2+ax+bx-48`. To find a and b, set up a system to be solved:

• 
  `a+b=13`

• 
  `ab=1(-48)=-48`

2.) List all such integer pairs that give product −48:

• 
  `-1,48`

• 
  `-2,24`

• 
  `-3,16`

• 
  `-4,12`

• 
  `-6,8`

3.) Calculate the sum for each pair:

• 
  `-1+48=47`

• 
  `-2+24=22`

• 
  `-3+16=13`

• 
  `-4+12=8`

• 
  `-6+8=2`

4.) The solution is the pair that gives sum 13:

• 
  `a=-3`

• 
  `b=16`

5.) Rewrite
`\bold{x^2+13x-48}`:

• 
  `x^2+13x-48=(x^2-3x)+(16x-48)`

6.) Factor out
`\bold{(x)}` in the first and 16 in the second group:

• 
  `x(x-3)+16(x-3)`

7.) Factor out common term
`\bold{x-3}` by using distributive property:

• 
  `(x-3)(x+16)`

So your answer is
`\bold{(x-3)(x+16)}` or
`\bold{(x+16)(x-3)}`.

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Answer 2

answer is (x+16)(x-3) I'm sure